The given integral is:

$\int_0^2 x^{-2} \, dx$

Step-by-step Solution:

1. Rewrite the integrand: $x^{-2} = \frac{1}{x^2}$

2. Use the power rule for integration: The integral of $x^n$ is: $\int x^n \, dx = \frac{x^{n+1}}{n+1}, \quad \text{for } n \neq -1.$

   Here, $n = -2$, so: $\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -x^{-1}.$

3. Evaluate the definite integral: Substitute the limits $0$ and $2$ into $-x^{-1}$: $\int_0^2 x^{-2} \, dx = \left[ -\frac{1}{x} \right]_0^2.$

   At $x = 2$: $-\frac{1}{2}.$

   At $x = 0$: The term $-\frac{1}{x}$ becomes undefined as $x \to 0^+$, meaning the integral diverges.

Conclusion:

This integral does not converge because $x^{-2}$ has an infinite value at the lower limit $x = 0$. The integral is divergent.

Would you like more details or further explanations?

Related Questions:

1. What is the meaning of divergence in integrals?
2. How can we handle improper integrals with infinity at one bound?
3. What is the difference between convergence and divergence in integrals?
4. How does the power of $x$ affect the behavior of improper integrals?
5. Can substitution methods be used for this integral?

Tip:

For improper integrals, always check the behavior of the function at the boundaries to determine whether the integral converges or diverges.